Question: Let $g$ be a vector-valued function defined by $g(t)=(2t^3,-2\sqrt{t+1})$. Find $g$ 's second derivative $g''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(6t^2,-\dfrac1{\sqrt{t+1}}\right)$ (Choice B) B $12t+\dfrac1{2\sqrt{(t+1)^3}}$ (Choice C) C $\left(12t,\dfrac1{2\sqrt{(t+1)^3}}\right)$ (Choice D) D $\left(6t,-\dfrac2{\sqrt{t+1}}\right)$
Solution: We are asked to find the second derivative of $g$. This means we need to differentiate $g$ twice. In other words, we differentiate $g$ once to find $g'$, and then differentiate $g'$ (which is a vector-valued function as well) to find $g''$. Recall that $g(t)=\left(2t^3,-2\sqrt{t+1}\right)$. Therefore, $g'(t)=\left(6t^2,-\dfrac1{\sqrt{t+1}}\right)$. Now let's differentiate $g'(t)=\left(6t^2,-\dfrac1{\sqrt{t+1)}}\right)$ to find $g''$. $g''(t)=\left(12t,\dfrac1{2\sqrt{(t+1)^3}}\right)$ In conclusion, $g''(t)=\left(12t,\dfrac1{2\sqrt{(t+1)^3}}\right)$.